Question

If the equation $4x^2+2\sqrt{3}xy+2y^2-1=0$ becomes $5x^2+y^2=1$, when the axes are rotated through an angle $\theta$, then $\theta$ is

• A. ${15}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

The correct option is B ${30}^o$

Given,

The equation

$4x^2+2\sqrt{3}xy+2y^2-1=0$

$\text{Let old coordinates}(x,y)$

$\text{new coordinates}(X,Y)$

$\because x=X\cos \theta -Y\sin \theta$

$y=X\sin \theta +Y\cos \theta$

Putting the value of (x) and (y) in equation (i)

$4(x\cos \theta -y\sin \theta {)}^2+2\sqrt{3}(x\cos \theta -y\sin \theta \left)\right(x\sin \theta +y\cos \theta )$

$+2(x\sin \theta +y\cos \theta {)}^2-1=0$

$\Rightarrow 4(x^2{\cos }^2\theta +y^2{\sin }^2\theta -2xy\cos \theta \sin \theta )+$

$2\sqrt{3}(x^2\cos \theta \sin \theta +x{\cos }^2\theta -xy{\sin }^2\theta -y^2\sin \theta \cos \theta )$

$+2(x^2{\sin }^2\theta +y^2{\cos }^2\theta +2xy\sin \theta \cos \theta )-1=0$

$\Rightarrow x^2(4{\cos }^2\theta +2{\sin }^2\theta +2\sqrt{3}\cos \theta \sin \theta )+y^2(4{\sin }^2\theta -2\sqrt{3}\sin \theta$

$+2{\cos }^2\theta )+xy(-8\sin \cot \theta +2\sqrt{3}{\cos }^2\theta -2\sqrt{3}{\sin }^2\theta$

$+4\sin \theta \cos \theta )-1=0$

equating this equation with $5x^2+y^2=1$,

$\therefore xy=0$

$\Rightarrow$ -. $-2\sqrt{3}({\sin }^2\theta -{\cos }^2\theta )+-4\sin \theta \cos \theta =0$

$\Rightarrow 2\sqrt{3}\cos 2\theta -2\sin 2\theta =D$

$\Rightarrow 2\times 2(\frac{\sqrt{3}}{2}\cos 2\theta -\frac{1}{2}\sin 2\theta )=0$

$\Rightarrow (\sin {60}^{\circ }\cos 2\theta -\cos {60}^{\circ }\sin 2\theta )=\theta$

$\Rightarrow \sin (60-2\theta )=0$

$\Rightarrow 60=2\theta [\because \sin 0^{\circ }=0]$

$\Rightarrow \theta ={30}^{\circ }$

Option (B) is correct